1.7E: Exercises for Section 1.7

In exercises 1 - 3, find the derivative $\dfrac{dy}{dx}$.

1) $y=\ln(2x)$

Answer

$\dfrac{dy}{dx} = \dfrac{1}{x}$

2) $y=\ln(2x+1)$

3) $y=\dfrac{1}{\ln x}$

Answer

$\dfrac{dy}{dx} = −\dfrac{1}{x(\ln x)^2}$

In exercises 4 - 5, find the indefinite integral.

4) $\displaystyle ∫\frac{dt}{3t}$

5) $\displaystyle ∫\frac{dx}{1+x}$

Answer

$\displaystyle ∫\frac{dx}{1+x} = \ln|x+1|+C$

In exercises 6 - 15, find the derivative $\dfrac{dy}{dx}$. (You can use a calculator to plot the function and the derivative to confirm that it is correct.)

6) [T] $y=\dfrac{\ln x}{x}$

7) [T] $y=x\ln x$

Answer

$\dfrac{dy}{dx} = \ln(x)+1$

8) [T] $y=\log_{10}x$

9) [T] $y=\ln(\sin x)$

Answer

$\dfrac{dy}{dx} = \cot x$

10) [T] $y=\ln(\ln x)$

11) [T] $y=7\ln(4x)$

Answer

$\dfrac{dy}{dx} = \frac{7}{x}$

12) [T] $y=\ln\big((4x)^7\big)$

13) [T] $y=\ln(\tan x)$

Answer

$\dfrac{dy}{dx} = \csc x\sec x$

14) [T] $y=\ln(\tan 3x)$

15) [T] $y=\ln(\cos^2x)$

Answer

$\dfrac{dy}{dx} = −2\tan x$

In exercises 16 - 25, find the definite or indefinite integral.

16) $\displaystyle ∫^1_0\frac{dx}{3+x}$

17) $\displaystyle ∫^1_0\frac{dt}{3+2t}$

Answer

$\displaystyle ∫^1_0\frac{dt}{3+2t} = \tfrac{1}{2}\ln\left(\tfrac{5}{3}\right)$

18) $\displaystyle ∫^2_0\frac{x}{x^2+1}\, dx$

19) $\displaystyle ∫^2_0\frac{x^3}{x^2+1}\,dx$

Answer

$\displaystyle ∫^2_0\frac{x^3}{x^2+1}\,dx = 2−\tfrac{1}{2}\ln(5)$

20) $\displaystyle ∫^e_2\frac{dx}{x\ln x}$

21) $\displaystyle ∫^e_2\frac{dx}{(x\ln x)^2}$

Answer

$\displaystyle ∫^e_2\frac{dx}{(x\ln x)^2} = \frac{1}{\ln(2)}−1$

22) $\displaystyle ∫\frac{\cos x}{\sin x}\, dx$

23) $\displaystyle ∫^{π/4}_0\tan x\,dx$

Answer

$\displaystyle ∫^{π/4}_0\tan x\,dx = \tfrac{1}{2}\ln(2)$

24) $\displaystyle ∫\cot(3x)\,dx$

25) $\displaystyle ∫\frac{(\ln x)^2}{x}\, dx$

Answer

$\displaystyle ∫\frac{(\ln x)^2}{x}\, dx = \tfrac{1}{3}(\ln x)^3$

In exercises 26 - 35, compute $\dfrac{dy}{dx}$ by differentiating $\ln y$.

26) $y=\sqrt{x^2+1}$

27) $y=\sqrt{x^2+1}\sqrt{x^2−1}$

Answer

$\dfrac{dy}{dx} = \dfrac{2x^3}{\sqrt{x^2+1}\sqrt{x^2−1}}$

28) $y=e^{\sin x}$

29) $y=x^{−1/x}$

Answer

$\dfrac{dy}{dx} = x^{−2−(1/x)}(\ln x−1)$

30) $y=e^{ex}$

31) $y=x^e$

Answer

$\dfrac{dy}{dx} = ex^{e−1}$

32) $y=x^{(ex)}$

33) $y=\sqrt{x}\sqrt[3]{x}\sqrt[6]{x}$

Answer

$\dfrac{dy}{dx} = 1$

34) $y=x^{−1/\ln x}$

35) $y=e^{−\ln x}$

Answer

$\dfrac{dy}{dx} = −\dfrac{1}{x^2}$

In exercises 36 - 40, evaluate by any method.

36) $\displaystyle ∫^{10}_5\dfrac{dt}{t}−∫^{10x}_{5x}\dfrac{dt}{t}$

37) $\displaystyle ∫^{e^π}_1\dfrac{dx}{x}+∫^{−1}_{−2}\dfrac{dx}{x}$

Answer

$π−\ln(2)$

38) $\dfrac{d}{dx}\left[\displaystyle ∫^1_x\dfrac{dt}{t}\right]$

39) $\dfrac{d}{dx}\left[\displaystyle ∫^{x^2}_x\dfrac{dt}{t}\right]$

Answer

$\dfrac{1}{x}$

40) $\dfrac{d}{dx}\Big[\ln(\sec x+\tan x)\Big]$

In exercises 41 - 44, use the function $\ln x$. If you are unable to find intersection points analytically, use a calculator.

41) Find the area of the region enclosed by $x=1$ and $y=5$ above $y=\ln x$.

Answer

$(e^5−6)\text{ units}^2$

42) [T] Find the arc length of $\ln x$ from $x=1$ to $x=2$.

43) Find the area between $\ln x$ and the $x$-axis from $x=1$ to $x=2$.

Answer

$\ln(4)−1) \text{ units}^2$

44) Find the volume of the shape created when rotating this curve from $x=1$ to $x=2$ around the $x$-axis, as pictured here.

45) [T] Find the surface area of the shape created when rotating the curve in the previous exercise from $x=1$ to $x=2$ around the $x$-axis.

Answer

$2.8656 \text{ units}^2$

If you are unable to find intersection points analytically in the following exercises, use a calculator.

46) Find the area of the hyperbolic quarter-circle enclosed by $x=2$ and $y=2$ above $y=1/x.$

47) [T] Find the arc length of $y=1/x$ from $x=1$ to $x=4$.

Answer

$s = 3.1502$ units

48) Find the area under $y=1/x$ and above the $x$-axis from $x=1$ to $x=4$.

In exercises 49 - 53, verify the derivatives and antiderivatives.

49) $\dfrac{d}{dx}\Big[\ln(x+\sqrt{x^2+1})\Big]=\dfrac{1}{\sqrt{1+x^2}}$

50) $\dfrac{d}{dx}\Big[\ln\left(\frac{x−a}{x+a}\right)\Big]=\dfrac{2a}{(x^2−a^2)}$

51) $\dfrac{d}{dx}\Big[\ln\left(\frac{1+\sqrt{1−x^2}}{x}\right)\Big]=−\dfrac{1}{x\sqrt{1−x^2}}$

52) $\dfrac{d}{dx}\Big[\ln(x+\sqrt{x^2−a^2})\Big]=\dfrac{1}{\sqrt{x^2−a^2}}$

53) $\displaystyle ∫\frac{dx}{x\ln(x)\ln(\ln x)}=\ln|\ln(\ln x)|+C$

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.